Question

(x+2)^2+(y-3)^2=25 is the equation of circle k(O). Find the equation of the line tangent to the circle at the x-intercepts.

Answer 1

Answer:
`y=-(4)/(3)x-8, y=(4)/(3)x-(8)/(3)`

Explanation:

The x-intercepts are when y=0.

`(x+2)^2 +3^2 =25\\\\(x+2)^2 =16\\\\x+2=\pm 4\\\\x=-2 \pm 4\\\\x=-6, 2`

`(x+2)^2 +(y-3)^2 =25\\\\2(x+2)+2(y-3) (dy)/(dx)=0\\\\(dy)/(dx)=-(x+2)/(y-3)`

If
`x=2, y=0, (dy)/(dx)=-(2+2)/(0-3)=(4)/(3)`.

• The equation of this tangent is
  `y=(4)/(3)x-(8)/(3)`.

If
`x=-6, y=0, (dy)/(dx)=-(4)/(3)`.

• The equation of the tangent is
  `y=-(4)/(3)x-8`